Correct. A binomial in the form $\displaystyle{{a}}^{{3}}-{{b}}^{{3}}\usepackage{cmbright}\Huge\$ can be factored as $\displaystyle{\left({a}-{b}\right)}{\left({{a}}^{{2}}+{a}{b}+{{b}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ .

$\displaystyle{125}{{m}}^{{3}}-{8}{{n}}^{{3}}={{\left({5}{m}\right)}}^{{3}}-{{\left({2}{n}\right)}}^{{3}}={\left({5}{m}-{2}{n}\right)}{\left({{\left({5}{m}\right)}}^{{2}}+{5}{m}\cdot{2}{n}+{{\left({2}{n}\right)}}^{{2}}\right)}={\left({5}{m}-{2}{n}\right)}{\left({25}{{m}}^{{2}}+{10}{m}{n}+{4}{{n}}^{{2}}\right)}\usepackage{cmbright}\Huge\$

The question mark stands for 10.